Let $K$ be a unital ring and $A$ a $K$ -algebra. Defining ``division'' requires special considerations when the algebras are non-associative so we introduce the definition in stages.

Associative division algebras

If $A$ is an associative algebra then we say $A$ is a division algebra if

(i)

$A$ is unital with identity $1$ . So for all $a\in A$ ,$$a1=1a=a$$

(ii)

Also every non-zero element of $A$ has an inverse. That is $a\in A$ , $a\neq 0$ , then there exists a $b\in A$ such that$$ab=1=ba$$ We denote $b$ by $a^{-1}$ and we may prove $a^{-1}$ is unique to $a$ .

The standard examples of associative division algebras are fields, which are commutative, and the non-split quaternion algebra: $\alpha,\beta\in K$ ,$$\left(\frac{\alpha,\beta}{K}\right)=\left\{ a_1 1+a_2 i+a_3 j+a_4 k : i^2=\alpha 1, j^2=\beta 1, k^2=-\alpha \beta 1, ij=k=-ji.\right\$$ where $x^2-\alpha$ and $x^2-\beta$ are irreducible over $K$ .

Non-associative division algebras

For non-associative algebras $A$ , the notion of an inverse is not immediate. We use $x.y$ for the product of $x,y\in A$ .

Invertible as endomorphisms: Let $a\in A$ . Then define $L_a:x\mapsto a.x$ and $R_a:x\mapsto x.a$ . As the product of $A$ is distributive, both $L_a$ an $R_a$ are additive endomorphisms of $A$ . If $L_a$ is invertible then we may call $a$ ``left invertible'' and similarly, when $R_a$ is invertible we may call $a$ ``right invertible'' and ``invertible'' if both $L_a$ and $R_a$ are invertible.

In this model of invertible, $A$ is a division algebra if, and only if, for each non-zero $a\in A$ , both $L_a$ and $R_a$ invertible. Equivalently: the equations $a.x=b$ and $y.a=b$ have unique solutions for nonzero $a,b\in A$ . However, $x$ and $y$ need not be equal.

A common method to produce non-associative division algebras of this sort is through Schur's Lemma.

Invertible in the product: In some instances, the notion of invertible via endomorphisms is not sufficient. Instead, assume $A$ has an identity, that is, an element $1\in A$ such that for all $a\in A$ ,$$1.a=a=a.1$$

Next if $a\in A$ , we say $a$ is invertible if there exists a $b\in A$ such that \begin{equation}\label{eq:inv} a.b=1=b.a \end{equation}and furthermore that for all $x\in A$ , \begin{equation}\label{eq:inv-non-a} b.(a.x)=x=(x.a).b. \end{equation}Evidently () can be inferred from (). This added assumption substitutes for the need of associativity in the proofs of uniqueness of inverses and in solving equations with non-associative products.

Proof. Let $x\in A$ . Then $xL_1=1.x=x=b.(a.x)=x L_a L_b$ . So $L_1=L_a L_b$ . As $L_1$ is the identity map, $L_a$ is injective and $L_b$ is surjective. As $A$ is finite dimensional, injective and surjective endomorphisms are bijective.

In this model, a non-associative algebra is a division algebra $A$ if it is unital and every non-zero element is invertible.

Alternative division algebras

The standard examples of non-associative division algebras are actually alternative alegbras, specfically, the composition algebras of fields, non-split quaternions and non-split octonions - only the latter are actually not associative. Invertible in the octonions is interpreted in the second stronger form.

This result is usually followed by two useful results which serve to omit the need to consider non-associative examples.